Question 484352: How do you solve this?
The dimensions of a rectangle are such that its length is 3 inches more than its width. If length were doubled and width decreased by 1, area would be increased by 176 inches squared. What are the length and width?
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! The dimensions of a rectangle are such that its length is 3 inches more than its width. If length were doubled and width decreased by 1, area would be increased by 176 inches squared. What are the length and width?
==========================================================================
Given: l = w + 3 [1]
The original area = A = lw
For the new area we can write
2l*(w-1) = A + 176
2(w+3)*(w-1) = A + 176
Substitute the value for l in [1] into the expressions for the two areas:
A = w(w+3) [2]
A + 176 = 2(w+3)(w-1) -> A = 2(w+3)(w-1) - 176
Since the LHS's are equal, we can equate the RHS's:
w(w+3) = 2(w+3)(w-1) - 176
Simplify and solve for w:
w^2 + 3w = 2w^2 + 4w - 6 - 176
w^2 + w - 182 = 0
w = (-1 +/- sqrt(1 + 4*182))/2
w = (-1 +/- 27)/2
Taking the positive solution, we have w = 13
Therefore l = 13 + 3 = 16
Ans: w = 13, l = 16
|
|
|
